tag:blogger.com,1999:blog-6933544261975483399.post981980443705994663..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 588: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, SumAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-83108701967509596482023-12-26T22:17:03.749-08:002023-12-26T22:17:03.749-08:00From problem 587, x=(a-c)/2=CH-AG=(CF-HF)-(AE+EG)=...From problem 587, x=(a-c)/2=CH-AG=(CF-HF)-(AE+EG)=(CD-e)-(AD+d)=2x-d-e<br />So, x=d+eMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57141535728953401922011-03-04T01:57:21.099-08:002011-03-04T01:57:21.099-08:00Using problem 587 and a similar method of proof, i...Using problem 587 and a similar method of proof, it is very easy.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80513182695479779962011-03-04T00:30:31.088-08:002011-03-04T00:30:31.088-08:00from 587
x = (a-c)/2, d = (a-b)/2, e = (b-c)/2
d...from 587<br />x = (a-c)/2, d = (a-b)/2, e = (b-c)/2<br />d + e = (a-b)/2 + (b-c)/2 = (a-c)/2 = xc .t . e. onoreply@blogger.com